A simple pendulum fixed inside a car has a time period of 4 seconds...

The initial time period of the simple pendulum is given by:

$T_1 = 2\pi \sqrt{\dfrac{L}{g}} = 4$

When the car accelerates with acceleration $a$, the effective acceleration due to gravity becomes:

$g_{eff} = \sqrt{g^2 + a^2} = g \left(1 + \dfrac{a^2}{g^2}\right)^{1/2}$

The new time period is given by:

$T_2 = 2\pi \sqrt{\dfrac{L}{g_{eff}}} = 3.99$

Taking the ratio of the time periods:

$\dfrac{T_2}{T_1} = \left( \dfrac{g}{g_{eff}} \right)^{1/2} = \left( 1 + \dfrac{a^2}{g^2} \right)^{-1/4}$

Since the change in time period is very small, $a$ is small compared to $g$. Using the binomial approximation $(1 + x)^n \approx 1 + nx$:

$\dfrac{T_2}{T_1} \approx 1 - \dfrac{1}{4} \dfrac{a^2}{g^2}$

Substituting the given values $T_1 = 4$ and $T_2 = 3.99$:

$\dfrac{3.99}{4} = 1 - \dfrac{1}{4} \dfrac{a^2}{g^2}$

$\dfrac{1}{4} \dfrac{a^2}{g^2} = 1 - \dfrac{3.99}{4}$

$\dfrac{1}{4} \dfrac{a^2}{g^2} = \dfrac{0.01}{4}$

$\dfrac{a^2}{g^2} = 0.01 = \dfrac{1}{100}$

Taking the square root on both sides:

$\dfrac{a}{g} = \dfrac{1}{10}$

$a = \dfrac{g}{10}$